# Symmetric group on finite set has Sylow subgroup of prime order

Suppose $n$ is a natural number greater than $1$. Let $S_n$ be the Symmetric group (?) of degree $n$. Then, there exists a prime $p$ such that $S_n$ has a $p$-Sylow subgroup (?) of order $p$, i.e., it is a Group of prime order (?).
1. Bertrand's postulate: The version we use states that for any $n \ge 2$, there exists a prime $p$ such that $p \le n < 2p$.
The symmetric group of degree $n$ has order $n!$. By fact (1), there exists a prime $p$ such that $n/2 < p \le n$. Thus, the largest power of $p$ dividing $n!$ is $p$. In particular, this means that the order of the $p$-Sylow subgroup (which exists by fact (2)) is $p$.